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### Solving a Quartic Diophantine Equation

```Date: 08/20/2005 at 13:09:51
From: Barukh
Subject: Quartic Diophantine Equation

I would like to know if there is a method of solving the following
equation in *rationals*:

x^4 + 14*x^2 + 1 = y^2.

I tried to re-write the equation as follows:

(x^2 + 7)^2 - 48 = y^2,

and then solve the equation z^2 - 48 = y^2.  There would not be a
problem if y and z were integers, but for rationals I don't know how
to proceed.

```

```
Date: 08/20/2005 at 22:02:12
From: Doctor Vogler
Subject: Re: Quartic Diophantine Equation

Hi Barukh,

Thanks for writing to Dr. Math.  That's a very good question.  When
you have an algebraic equation (a polynomial) in two variables (this
is an algebraic curve), and you want to find its rational solutions,
this is what you do:  First, you determine the genus of the curve.
Computing the genus is slightly complicated, and more involved than is
usually necessary.  So I'll tell you a few special cases and then only
go into greater detail if you ask.  Every curve has a nonnegative
genus, and the largest the genus can be is (d-1)(d-2)/2, where d is
the degree of the polynomial.  So any polynomial in two variables with
degree 1 or 2 has genus zero.  Most polynomials of degree three have
genus 1 (but not all; the singular ones have genus 0).  A curve of the
form

y^2 = polynomial in x,

where the polynomial on the right has no repeated roots, has genus one
less than half of the degree of the polynomial in x, rounded up.
That's what you have.

When you have a curve of genus zero, you can write out a parametric
formula for all rational solutions.  The most common and the simplest
case of this is when you have a degree-two polynomial.  To see how to
solve this case, refer to this answer from our archives.

Rational Solutions to Two Variable Quadratic Equation
http://mathforum.org/library/drmath/view/65319.html

For other curves of genus zero, refer to this paper by Mark Van Hoeij:

A curve of genus one is called an "elliptic curve."  Most of the study
of elliptic curves is done by analyzing a degree-three equation of a
particular form, called Weierstrass form.  Van Hoeij also gives an
algorithm to convert any genus-one curve into Weierstrass form:

but for the two most common cases, including yours, I would refer you
to the formulas in Connell's manual on elliptic curves, available at

http://www.math.mcgill.ca/connell/public/ECH1/

The formulas you need are in chapter 1, section 1.2.  For your curve,

x^4 + 14*x^2 + 1 = y^2,

it says to define the new variables

u = (2y + 2)/x^2
v = (4y + 28x^2 + 4)/x^3

because these can give you the x-y values back by the formulas

x = (2u + 28)/v
y = -1 + (1/2)ux^2,

and the equation when converted into u and v becomes a cubic in
Weierstrass form.  Substitute the formulas for y and then x into your
equation, simplify, and you get (as long as x is not zero and u is not
-14)

v^2 = u^3 + 14u^2 - 4u - 56
= (u - 2)(u + 2)(u + 14).

Now you can analyze this elliptic curve.  Its Weierstrass parameters
are 0, 14, 0, -4, -56 (which will be needed later).  It takes a lot of
work from the study of elliptic curves to prove that the rational
points on an elliptic curve form an abelian group (like you studied in
abstract algebra).  The law for "adding" points is a geometric line
construction described in some detail (with examples) in this answer
in our archives:

Cubic Diophantine Equation in Three Variables
http://mathforum.org/library/drmath/view/66650.html

So then the only question that remains is the form of the group, and
generators.  To make a long story less long, the rational points of
any elliptic curve form a group of the form

T x Z^r,

where T is a finite group (the "torsion" points, all points with
finite order) and Z^r is the Cartesian product of r copies of Z (the
integers).  The number r is called the "rank" of the elliptic curve.
Furthermore, there are only a handful of possibilities for the group
T, which is either cyclic of rather small order, or is the Cartesian
product of a cyclic group of order two with another cyclic group of
rather small order.  It is actually not very difficult to determine
the torsion group of an elliptic curve, but without going into the
mathematics of it, I will simply suggest asking the math program GNU

http://pari.math.u-bordeaux.fr/

You can ask Pari

e = ellinit([0,14,0,-4,-56])

to initialize the elliptic curve, and then ask

elltors(e)

for the torsion group.  It will tell you

[8, [4, 2], [[10, 48], [-2, 0]]]

meaning that the torsion group has order 8, with the form

Z/4Z x Z/2Z

and with generators

(u, v) = (10, 48)

and

(u, v) = (-2, 0).

The other six points in the torsion group can be found by combining
these:  There is the identity (zero) point, also known as the "point
at infinity," and then there are

which Pari will tell you are

[2, 0]
[10, -48]
[-6, 16]
[-14, 0]
[-6, -16].

That done, we would like to find out the number r, the rank of the
curve.  This is, in fact, a much harder problem, and there is (as far
as I know) only one program that does this.  It is called MWRank, by
John Cremona, and can be downloaded from his web page at

http://www.maths.nott.ac.uk/personal/jec/

along with a paper describing how it works.  You give it

0 14 0 -4 -56

and it gives you a lot more information than you know what to do with,
but finally concludes that the rank is zero.

That means that the eight torsion points are the only rational points
on your curve, namely

[10, 48]
[-2, 0]
[2, 0]
[10, -48]
[-6, 16]
[-14, 0]
[-6, -16],

and the point at infinity.  When we convert these back into x-y
coordinates, we get

(1, 4),
the point at infinity,
the point at infinity,
(-1, 4),
(1, -4),
an invalid point,
(-1, -4),
and another invalid point.

It takes some fancy work to show that the two invalid points should
actually go to the obvious solutions

(0, 1)  and  (0, -1).

So we conclude that these six points are the only solutions to your
equation with rational x and y values.  It is interesting that the
only rational solutions are all integer solutions.  In fact, this is
common when the rank of the curve is zero, but not when the rank is
positive.

Given such a simple conclusion, we might want to see if we can prove
this in a more elementary fashion.  And perhaps we can!  Or perhaps not.

back and show me what you have been able to do, and I will try to
offer further suggestions.

- Doctor Vogler, The Math Forum
http://mathforum.org/dr.math/

```

```
Date: 08/21/2005 at 05:32:48
From: Barukh
Subject: Quartic Diophantine Equation

Hi Doctor Vogler,

Thank you very much for such a detailed explanation.  I didn't expect
to get such a thorough treatment of the problem.  It is just
excellent!

I was familiar with the elliptic curves method a bit, but didn't know
it is suitable for other equations except Weierstrass form.  I will
definitely look at the links you sent me, and try to perform all the
needed steps.

If you are curious to know how I arrived at this equation, I can tell
you that I tried to find rational solutions to a pair of quadratic DE:

r^2 + t^2 = u^2, r^2 + (2t)^2 = v^2.

Another QDE that arises from this pair has even simpler form:

x^4 - x^2 + 1 = y^2.

Best regards,

Barukh.
```
Associated Topics:
College Number Theory

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